1. Introduction. B C

ASIAN J. MATH. c© 2004 International PressVol. 8, No. 2, pp. 233–258, June 2004 002

MAPS BETWEEN Bn AND BN WITH GEOMETRIC RANKk0 ≤ n− 2 AND MINIMUM N ∗

SHANYU JI† AND DEKANG XU†

Dedicated to Professor Yum-Tong Siu on the Occasion of his 60th Birthday

1. Introduction. Let Bn = {z ∈ Cn : |z| < 1} be the unit ball in Cn. The prob-lem of classifying proper holomorphic mappings between Bn and BN has attractedconsiderable attention (see [Fo 1992] [DA 1988] [DA 1993] [W 1979] [H 1999][HJ 2001]for extensive references) since the work of Poincare [P 1907][T 1962] and Alexan-der [A 1977]. Let us denote by Prop(Bn,BN ) the collection of proper holomorphicmappings from Bn to BN . It is known [A 1977] that any map F ∈ Prop(Bn,Bn)must be biholomorphic and must be equivalent to the identity map. Here we say thatf, g ∈ Prop(Bn,BN ) are equivalent if there are automorphisms σ ∈ Aut(Bn) andτ ∈ Aut(BN ) such that f = τ ◦ g ◦ σ. For general N > n, the discovery of inner func-tions indicates that Prop(Bn,BN ) is too complicated to be classified. Hence we mayfocus on Rat(Bn,BN ), the collection of all rational proper holomorphic mappingsfrom Bn to BN . We first recall the following results:

Theorem 1.0. Let 2 ≤ n ≤ N .

(1) [We 1979][Fa 1986] When N < 2n − 1, Rat(Bn,BN ) has only one equiva-lent class.(2a) [Fa 1982] When N = 2n−1 and n = 2, Rat(B2,B3) has four equivalent classes.(2b) [HJ 2001] When N = 2n − 1 and n > 2, Rat(Bn,B2n−1) has exactly twoequivalent classes. One is the linear map and another one is Whitney map.(3) [DA 1988] When N = 2n, Rat(Bn,B2n) has infinitely many equivalent classes.In particular, {Ft(z1 · · · , zn) = (z1, · · · , zn−1, cos(t)zn, sin(t)znz) : t ∈ (0, π/2)} is afamily of mutually inequivalent polynomial proper embeddings.

However a puzzle remains: Why is the case of n = 2 in Theorem 1.0 (2a) morecomplicated than the one of n ≥ 3 in Theorem 1.0(2b)?

This puzzle can be solved by the following new formulation, which is cruciallybased on a notion, geometric rank, introduced recently by Huang [H 2003]. For any2 ≤ n ≤ N , any F ∈ Rat(Bn,BN ) can be associated an invariant integer κ0 ∈{1, 2, ..., n− 1}, called its geometric rank (see § 2 for the definition). It is known thatfor any F ∈ Rat(Bn,BN ), its geometric rank κ0 = 0 if and only if F is equivalentto the linear map ([H 1999, Theorem 4.2] cf. [HJ 2001, Propostion 2.2]). Therefore,to study maps in Rat(Bn,BN ), it is sufficient to study maps with geometric rankκ0 ≥ 1.

If F ∈ Rat(Bn,BN ) with the geometric rank κ0, it is known [H 2003, lemma3.2] that N ≥ n + (2n−κ0−1)κ0

2 must hold, namely, the least dimension of the targetspace is n+ (2n−κ0−1)κ0

2 . Therefore, to understand the simplest case, given an integerκ0 ∈ {1, ..., n − 1}, we are interested in studying maps F ∈ Rat(Bn,BN ) with the

∗Received June 25, 2003; accepted for publication October 20, 2003.†Department of Mathematics, University of Houston, Houston, TX 77204-3008, USA

([email protected]; [email protected]).

233

234 S. JI AND D. XU

geometric rank κ0 and with the minimum dimension of the target space:

F : Bn → BN , N = n+(2n− κ0 − 1)κ0

2. (1)

When κ0 = 1, (1) becomes F : Bn → B2n−1, which is the case covered by Theorem1.0 (2a) and (2b).

It is also known from a recent deep and important result by Huang [H 2003]that there is a significant difference between the case 1 ≤ κ0 ≤ n − 2 and the caseκ0 = n − 1. More precisely, when 1 ≤ κ0 ≤ n − 2, the maps F have so-calledsemi-linear property while the maps with κ0 = n − 1 may not have such property.This gives a philosophy that maps F with κ0 = n − 1 are comparatively much morecomplicated than the ones with 1 ≤ κ0 ≤ n− 2. From this philosophy, the followingtwo problems are naturally formulated.

Problem A. Study and classify maps F ∈ Rat(Bn,BN ) with N = n+ (2n−κ0−1)κ02

and 1 ≤ κ0 ≤ n− 2.

Problem B. Study and classify maps F ∈ Rat(Bn,BN ) with N = n+ (2n−κ0−1)κ02

and κ0 = n− 1.

When κ0 = 1, Problem A is solved in Theorem 1.0 (2b), and beyond this case,the next simplest unsolved case is Rat(B4,B9) with κ0 = 2. When n = 2, Problem Bis solved in Theorem 1.0 (2a), and beyond this case, the next simplest unsolved caseis Rat(B3,B6) with κ0 = 2.

In fact, the formulation of Problems A and B explains why Rat(B2,B3) is morecomplicated than Rat(Bn,B2n−1) with n ≥ 3: Each of Theorem 1.0 (2a) and (2b) isan initial case of Problem A and Problem B.

In this paper, we study Problem A and we first estimate the degree of such mapsF . As the main result, we investigate Problem A by studying maps F ∈ Rat(B4,B9)with κ0 = 2 and deg(F ) = 2.

Theorem 1.1. Let F ∈ Rat(Bn,BN ) with geometric rank κ0, 1 ≤ κ0 ≤ n − 2,and with N = n+ (2n−κ0−1)κ0

2 . Then deg(F ) ≤ κ0 + 2.

Theorem 1.2. Let F ∈ Rat(B4,B9) with the geometric rank 2 and with deg(F ) =2. Then F is equivalent to Whitney map W4,2 of rank 2.

The paper is organized as follows. We first prove Theorem 1.1, by using the sametechnique in the proof of Lemma 5.2 in [HJ 2001]. Here we would like to mentiona conjecture by D’Angelo [DA 1993, p.189] which is open: if F ∈ Rat(Bn,BN ),then deg(F ) ≤ 2N − 3. Next we introduce the definition of Whitney map of rankκ0 ( see (9)) and prove a criterion for such maps. In Section 5, we determine theform of F (z, 0), in which the semi-linearity property of F by Huang [H 2003] willbe crucially used. In Section 6, we further determine the form of F (z, w). Theorem6.1 tells us what F looks like when F ∈ Rat(H4,H9) satisfies the normalizationcondition in Theorem 2.2: F has three complex parameters b(11)1001, b

(13)1001, E0001 and one

real parameter µ2 ≥ 1 that are related by certain equations. To prove such mapF is equivalent to Whitney map, one key step is to change the parameter µ2 into 1.However, the difficulty is that such µ2 is invariant under any equivalent change that fixthe origin (see (4)(5)). If we consider F ∗∗∗

p , which is equivalent to F (see Theorem 2.2),the calculation of µ2,F (p) is too complicated to be handled. Our idea is to calculate

MAPS BETWEEN Bn AND BN WITH k0 ≤ n − 2 AND MINIMUM N 235

only the linear part of µ2,F (p), which dramatically reduces our computation. As aresult, we see that µ2,F (p) decreases when p moves along a certain direction. Thenwe are able to show that µ2,F (p) can reach the minimum and hence it must be 1 andthe resulting map F is exactly Whitney map.

Acknowledgments. We wish to thank Xiaojun Huang for helpful discussionsand valuable suggestion for our work in this paper. We also thank the referees whogave very useful suggestions and comments.

2. Preliminaries. Let Hn := {(z, w) ∈ Cn−1 × C, Im(w) > |z|2} be the Siegelupper-half space in Cn where n ≥ 2. By using the Cayley transformation:

ρn : Hn → Bn, ρn(z, w) =(

2z1 − iw

,1 + iw

1 − iw

), ρ−1

n (z∗, w∗) =(

z∗

w∗ + 1,1i· w

∗ − 1w∗ + 1

),

Hn is biholomorphic equivalent to the unit ball Bn and ∂Hn is equivalent to the unitsphere ∂Bn. For any F ∈ Rat(Bn,BN ), ρ−1

N ◦ F ◦ ρn ∈ Rat(Hn,HN ). Then we canidentify a map F ∈ Rat(Bn,BN ) with the one in Rat(Hn,HN ), and we shall stilldenote it as F for simplicity. By the work of Cima and Suffridge[CS 1990], F extendsholomorphically up to the boundary. Hence the map F induces a non-constant CRmapping from ∂Hn to ∂HN . As before, we also denote it as F .

Write Lj = 2izj∂

∂w + ∂∂zj

for j = 1, · · · , n − 1 and T = ∂∂u with w = u + iv.

{L1, · · · , Ln−1} forms a basis for the complex tangent bundle T (1,0)∂Hn, and T isthe tangent vector field of ∂Hn transversal to T (1,0)∂Hn ∪ T (0,1)∂Hn. ∂Hn canbe parameterized by (z, z, u) through the map (z, z, u) → (z, u + i|z|2). Assign theweight of z and u to be 1 and 2 respectively. If m is a non-negative integer, a functionh defined over a neighborhood U of 0 in ∂Hn is said to be of quantity owt(m) ifh(tz,tz,t2u)

|t|m → 0 uniformly for (z, u) on any compact subset of U as t(∈ R) → 0. Forthis case, we write h = owt(m).

Let F = (f, φ, g) = (f , g) = (f1, · · · , fn−1, φ1, · · · , φN−n, g) be a map inRat(Hn,HN ). For any p = (z0, w0) ∈ ∂Hn, we consider automorphisms σ0

(z0,w0)∈

Aut(Hn) and τF(z0,w0)

∈ Aut(HN ) given by

σ0(z0,w0)

(z, w) = (z + z0, w + w0 + 2i〈z, z0〉),

τF(z0,w0)

(z∗, w∗) = (z∗ − f(z0, w0), w∗ − g(z0, w0) − 2i〈z∗, f(z0, w0)〉).

Then Fp = τFp ◦ F ◦ σ0

p ∈ Rat(Hn,HN ) is equivalent to F with Fp(0) = 0. By thework of Huang [H 1999], Fp is equivalent to another new map F ∗∗

p = (f∗∗p , φ∗∗p , g∗∗p )

that satisfies the following normalization condition.

Theorem 2.1. [H 1999, Lemma 5.3] Let F be a C2-smooth CR map from aconnected open subset M ⊂ ∂Hn into ∂HN with N ≥ n ≥ 2. Then for each p ∈ M ,F ∗∗

p = (f, φ, g) satisfies the normalization condition:

fp = z +i

2a(1)

p (z)w + owt(3), φp = φ(2)p (z) + owt(2), gp = w + owt(4) (2)

236 S. JI AND D. XU

with 〈z, a(1)p (z)〉|z|2 = |φ(2)

p (z)|2, where we denote by h(j)(z) a polynomial of z withhomogeneous degree j.

Here a(1)p (z) = zA(p) where A(p) is a certain (n − 1) × (n − 1) semi-positive

Hermitian matrix. The rank of A(p) is said to be the geometric rank of F at thepoint of p. We denote it by RkF (p). We define the geometric rank of F to beκ0 = maxp∈∂HnRkF (p). Notice that 0 ≤ κ0 ≤ n− 1.

Theorem 2.2. [H 2003, Lemma 3.2, 3.3, Corollary 4.2, 5.2, (3.6.4) and Claim4.4] Let F be a C3-smooth CR map from a connected open subset M ⊂ ∂Hn into∂HN with F (0) = 0, 1 ≤ κ0 = RkF (0) ≤ n − 2 and N = n + (2n−κ0−1)κ0

2 . Then for∀p(≈ 0) ∈ M , Fp is equivalent to another map F ∗∗∗

p , still denote it by (f, φ, g), from∂Hn to ∂HN , with the following conditions:

fj = zj + iµj

2 zjw + owt(3),fj = zj + owt(3), ∂2fj

∂w2 (0) = 0, 1 ≤ j ≤ n− 1;φjl = µjlzjzl + owt(2), ∂2φjl

∂zk∂w (0) = 0 for k > κ0,∂2φjl

∂w2 (0) = 0, ∀(j, l) ∈ S0;g = w + owt(5), g(0, w) = w + o(w3),

(3)

where S0 = {(jl), 1 ≤ j ≤ κ0, 1 ≤ l ≤ n − 1, j ≤ l}, µj ≥ µ1 = 1 for j ≤ κ0

and µj = 0 for j > κ0; µjl =√µj + µl for 1 ≤ j ≤ κ0, 1 ≤ l ≤ n − 1, j = l and

µjj = √µj for 1 ≤ j ≤ κ0.

We notice that S0 is the finite index set of {φjl} and |S0| = (2n−κ0−1)κ02 . Also,

by [H 2003, Corollary 5.2], the set {p ∈ ∂Hn, RkF (p) = κ0} is an open dense subsetof M . Therefor the assumption that F (0) = 0, κ0 = RkF (0) holds for almost p ∈M .

For any rational holomorphic map H = (P1,··· ,Pm)Q on Cn, where Pj , Q are

holomorphic polynomials and (P1, · · · , Pm, Q) = 1, the degree of H is defined to bedeg(H) = max{deg(Pj), 1 ≤ j ≤ m, deg(Q)}.

Lemma 2.3. [HJ 2001, Lemma 5.3 and 5.4] Let F ∈ Rat(Hn,HN ). Ifdeg(Fp(z, 0)) ≤ l for any p in an open subset of ∂Hn, then deg(F ) ≤ l.

Consider σ ∈ Aut0(Hn) and τ∗ ∈ Aut0(HN ) given by

σ =(λ(z + aw) · U, λ2w)

q(z, w), (4)

where q(z, w) = 1 − 2i〈a, z〉 + (r − i|a|2)w, λ > 0, r ∈ R, a ∈ Cn−1 and U is an(n− 1) × (n− 1) unitary matrix, and

τ∗(z∗, w∗) =λ∗(z∗ + a∗w∗) · U∗, λ∗2w∗

q∗(z∗, w∗), (5)

where q∗(z∗, w∗) = 1 − 2i〈a∗, z∗〉 + (r∗ − i|a∗|2)w∗, λ∗ > 0, r∗ ∈ R, a∗ ∈ CN−1 andU∗ is an (N − 1) × (N − 1) unitary matrix.

Lemma 2.4. ([H 2003, Lemma 2.2]) Let F = (f, φ, g) and F ∗ = (f∗, φ∗, g∗) beC2 CR map from a neighborhood of 0 in ∂Hn into ∂HN (N ≥ n > 1) such that both


satisfy the normalization condition (2). Suppose that F ∗ = τ∗ ◦F ◦ σ where σ and τ∗

are as in (4) and (5) respectively. Then it holds that

λ∗ = λ−1, a∗1 = −λ−1a · U, a∗2 = 0, r∗ = −λ−2r, U∗ =(U−1 0

0 U∗22

), (6)

where a∗ = (a∗1, a∗2) with a∗1 its first (n− 1) components, U∗

22 is an (N − n)× (N − n)unitary matrix. Conversely, suppose τ∗ and σ, given in (4) and (5) respectively,are related by (6). Suppose that F satisfies (2). Then F ∗ := τ∗◦F ◦σ also satisfies (2).

3. Proof of Theorem 1.1 . Assume that F satisfies (3). Let Lj = 2iζj ∂∂w + ∂

∂zj

be the complexfication of Lj . We apply Lj ,LkLj(k ≤ κ0, k ≤ j) to the equation:g(z,w)−g(ζ,η)

2i = f(z, w)f(ζ, η)+φ(z, w)φ(ζ, η) for any w−η = 2i(z · ζ). Let (z, w) = 0,η = 0. Then we get

f(ζ, 0)t =(

I 0−B−1A B−1

)(ζ

t

0

)=

(ζ

t

−B−1Aζt

), ∀ζ ∈ Cn−1, (7)

where

A =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

L1L1(f1) L1L1(f2) · · · L1L1(fn−1)L1L2(f1) L1L2(f2) · · · L1L2(fn−1)

· · · · · · · · · · · ·L1Ln−1(f1) L1Ln−1(f2) · · · L1Ln−1(fn−1)L2L2(f1) L2L2(f2) · · · L2L2(fn−1)

· · · · · · · · · · · ·L2Ln−1(f1) L2Ln−1(f2) · · · L2Ln−1(fn−1)

· · · · · · · · · · · ·Lκ0Ln−1(f1) Lκ0Ln−1(f2) · · · Lκ0Ln−1(fn−1)

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

∣∣∣∣(0,0,ζ,0)

=

⎛⎜⎜⎜⎝C1

C2

...Cκ0

⎞⎟⎟⎟⎠

and Ci are (n− i) × (n− 1) matrices with the following forms:

C1 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

−2ζ1 0 0 · · · 0 0 · · · 0−ζ2 −µ2ζ1 0 · · · 0 0 · · · 0−ζ3 0 −µ3ζ1 · · · 0 0 · · · 0

......

......

......

......

−ζκ00 0 · · · −µκ0ζ1 0 · · · 0

−ζκ0+1 0 0 · · · 0 0 · · · 0...

......

......

......

...−ζn−1 0 0 · · · 0 0 · · · 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠,

C2 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 −2µ2ζ2 0 · · · 0 0 · · · 00 −µ2ζ3 −µ3ζ2 · · · 0 0 · · · 0...

......

......

......

...0 −µ2ζκ0

0 · · · −µκ0ζ2 0 · · · 00 −µ2ζκ0+1 0 · · · 0 0 · · · 0...

......

......

......

...0 −µ2ζn−1 0 · · · 0 0 · · · 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠,

238 S. JI AND D. XU

· · · , Cκ0 =

⎛⎜⎜⎜⎝0 0 0 · · · −2µκ0ζκ0

0 · · · 00 0 0 · · · −µκ0ζκ0+1 0 · · · 0...

......

......

......

...0 0 0 · · · −µκ0ζn−1 0 · · · 0

⎞⎟⎟⎟⎠ ,

and B is equal to

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

L1L1(φ11) · · · L1L1(φ1(n−1)) L1L1(φ22) · · · L1L1(φκ0(n−1))L1L2(φ11) · · · L1L2(φ1(n−1)) L1L2(φ22) · · · L1L2(φκ0(n−1))

· · · · · · · · · · · · · · · · · ·L1Ln−1(φ11) · · · L1Ln−1(φ1(n−1)) L1Ln−1(φ22) · · · L1Ln−1(φκ0(n−1))L2L2(φ11) · · · L2L2(φ1(n−1)) L2L2(φ22) · · · L2L2(φκ0(n−1))

· · · · · · · · · · · · · · · · · ·L2Ln−1(φ11) · · · L2Ln−1(φ1(n−1)) L2Ln−1(φ22) · · · L2Ln−1(φκ0(n−1))

· · · · · · · · · · · · · · · · · ·Lκ0Ln−1(φ11) · · · Lκ0Ln−1(φ1(n−1)) Lκ0Ln−1(φ22) · · · Lκ0Ln−1(φκ0(n−1))

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

∣∣∣∣(0,0,ζ,0)

.

Hence Aζt=

⎛⎜⎜⎜⎝D1

D2

...Dκ0

⎞⎟⎟⎟⎠ with

D1 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

−2ζ2

1

−(1 + µ2)ζ1ζ2

−(1 + µ3)ζ1ζ3...

−(1 + µκ0)ζ1ζκ0

−ζ1ζκ0+1...

−ζ1ζn−1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠, D2 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

−2µ2ζ2

2

−(µ2 + µ3)ζ2ζ3...

−(µ2 + µκ0)ζ2ζκ0

−µ2ζ2ζκ0+1...

−µ2ζ2ζn−1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠, ..., Dκ0 =

⎛⎜⎜⎜⎝−2µκ0ζ

2

κ0

−µκ0ζκ0ζκ0+1

...−µκ0ζκ0

ζn−1

⎞⎟⎟⎟⎠ .

Notice that LiLj(φst(0, 0)) = 2iζi∂2φst(0,0)

∂w∂zj+2iζj

∂2φst(0,0)∂w∂zi

+ ∂2φst(0,0)∂zi∂zj

from (3). If we

denote bkij to be ∂2φij

∂w∂zk|0, then we get

LiLi(φii) = 2√µi + 4iζib

iii; LiLi(φij) = 4iζib

iij , i = j;

LiLj(φij) =√µi + µj + 2iζib

jij + 2iζib

iij , for i < j ≤ κ0;

LiLj(φst) = 2iζibjst + 2iζib

ist, for (ij) = (st), i < j ≤ κ0;

LiLj(φij) =√µi + µj + 2iζjb

iij , for j > κ0;

LiLj(φst) = 2iζjbist, for j > κ0.


Recalling µ1 = 1, µj = 0 for j > κ0, we can write B = D + B with

D =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

2 0 · · · 0 0 · · · 0 0 0 0 00

√1 + µ2 · · · 0 0 · · · 0 0 0 0 0

......

......

......

......

......

...0 0 · · · √

1 + µκ0 0 · · · 0 0 0 0 00 0 · · · 0 1 · · · 0 0 0 0 0...

......

......

......

......

......

0 0 · · · 0 0 · · · 1 0 0 0 00 0 · · · 0 0 · · · 0 2

√µ2 0 0 0

0 0 · · · 0 0 · · · 0 0√µ2 + µ3 0 0

......

......

......

......

......

...0 0 · · · 0 0 · · · 0 0 0 · · · √

µκ0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

,

B =

⎛⎜⎝ E1

...Eκ0

⎞⎟⎠⎛⎜⎝b

111 b112 · · · b11(n−1) b122 · · · b1κ0(n−1)

......

......

......

...bκ011 bκ0

12 · · · bκ01(n−1) bκ0

22 · · · bκ0κ0(n−1)

⎞⎟⎠ ,

where

E1 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

4iζ1 0 0 · · · 02iζ2 2iζ1 0 · · · 0

......

......

...2iζκ0

0 0 · · · 2iζ1

2iζκ0+1 0 · · · · · · · · ·...

......

......

2iζn−1 0 0 · · · 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠, E2 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 4iζ2 0 · · · 00 2iζ3 2iζ2 · · · 0...

......

......

0 2iζκ00 · · · 2iζ2

0 2iζκ0+1 0 · · · 0...

......

......

0 2iζn−1 0 · · · 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠,

. . . , Eκ0 =

⎛⎜⎜⎜⎝0 0 0 · · · 4iζκ0

0 0 0 · · · 2iζκ0+1...

......

......

0 0 0 · · · 2iζn−1

⎞⎟⎟⎟⎠ .

Denote B∗ = D−1B = 2i�ζ ·�b where

D−1

⎛⎜⎝ E1

...Eκ0

⎞⎟⎠ = 2i�ζ, and �b =

⎛⎜⎝b111 b112 · · · b11(n−1) b122 · · · b1κ0(n−1)

......

......

......

...bκ011 bκ0

12 · · · bκ01(n−1) bκ0

22 · · · bκ0κ0(n−1)

⎞⎟⎠ .

Then (B∗)2 = (2i)2�ζ · (�b · �ζ) ·�b, · · · , (B∗)k = (2i)k�ζ · (�b · �ζ)k−1 ·�b, · · · . Hence

−B−1Aζt= −(I +B∗)−1D−1Aζ

t= −(I +

∑∞j=1(−1)jB∗j)D−1Aζ

t

= −[I − 2i�ζ(∑∞

j=1(−1)j−1(2i)j−1(�b · �ζ)j−1) ·�b]D−1Aζ

t

= −[I − 2i�ζ(I + 2i�b · �ζ)−1 ·�b]D−1Aζ

t.

(8)

240 S. JI AND D. XU

Notice that �b · �ζ is a κ0 × κ0 matrix. Each entry of this matrix is a polynomialof ζ with degree 1. We also notice that D−1Aζ

tis a vector of polynomial about ζ

with degree 2, we conclude deg φ ≤ κ0 + 2. If we let z = w = η = 0 in g(z,w)−g(ξ,η)2i =

f(z, w)f(ξ, η), we get g(ζ, 0) = 0. Hence deg(F ∗∗∗p ) ≤ κ0 + 2 for any p in ∂Hn that is

closed to 0. By Lemma 2.3, deg(F ) ≤ κ0 +2. The proof of Theorem 1.1 is completed.

4. Whitney Maps of Rank κ0. Let 1 ≤ κ0 ≤ n− 1, N = n+ (2n−κ0−1)κ02 . A

map Wn,κ0 = (Γ1, · · · ,Γκ0+1) in Rat(Bn,BN ) is called Whitney map of rank κ0 if itis of the following form:

Γ1 = (z21 ,√

2z1z2, · · · ,√

2z1zκ0 , z1zκ0+1, · · · , z1zn−1, z1w),Γ2 = (z2

2 ,√

2z2z3, · · · ,√

2z2zκ0 , z2zκ0+1, · · · , z2zn−1, z2w),...Γκ0 = (z2

κ0, zκ0zκ0+1, · · · , zκ0zn−1, zκ0w),

Γκ0+1 = (zκ0+1, · · · , zn−1, w).

(9)

Notice that when κ0 = 1, Γ1 = (z21 , z1z2, ..., z1zn−1, z1w) and Γ2 = (z2, · · · , zn−1, w)

give the classical Whitney map. By Cayley transformation, Wn,κ0 can be identifiedas a map in Rat(Hn,HN ). As an example, W4.2 ∈ Rat(H4,H9) is of the form

f1 = z1+i4 z1w

1− i4 w

, f2 = z2+i4 z2w

1 i4 w

, f3 = z3− i4 z3w

1− i4 w

,

φ11 = z21

1− i4 w, φ12 =

√2z1z2

1− i4 w

, φ13 = z1z31− i

4 w, φ22 = z2

21− i

4 w, φ23 = z2z3

1− i4 w, g = w.

(10)We want to prove a criterion for Whitney map which will be used to prove Theorem1.2.

Theorem 4.1. Let F ∈ Rat(Hn,HN ) with the geometric rank κ0, 1 ≤ κ0 ≤ n−2,and with N = n+ (2n−κ0−1)κ0

2 . Then F is equivalent to Wn,κ0 if and only if deg(F ) = 2

and F is equivalent to another map F = (f, φ, g) that satisfies (3) and ∂2φjl

∂zk∂w (0, 0) = 0for all j, l and k.

Proof. It suffices to show that if F satisfies (3) and ∂2φjl

∂zk∂w (0, 0) = 0 for all j, land k, then F is equivalent to Wn,κ0 .

Step 1. Determine F (z, 0). As we did in §3, apply Lj ,LkLj(k ≤ κ0, k ≤ j) to

g(z, w) − g(ζ, η)2i

= f(z, w)f(ζ, η) + φ(z, w)φ(ζ, η), (11)

for any w − η = 2i(z · ζ).By (7), we have

⎛⎜⎜⎜⎝f1(ζ, 0)

...fn−1(ζ, 0)φ(ζ, 0)

⎞⎟⎟⎟⎠ =(

I 0−B−1A B−1

)⎛⎜⎜⎜⎝ζ1...

ζn−1

0

⎞⎟⎟⎟⎠ ,


where A is as in Section 3, and

B =

⎛⎜⎜⎜⎝D(n−1)×|S0|D(n−2)×|S0|

...D(n−κ0)×|S0|

⎞⎟⎟⎟⎠is an |S0| × |S0| = (2n−κ0−1)κ0

2 × (2n−κ0−1)κ02 diagonal matrix with

D(n−1)×|S0| =

⎛⎜⎜⎜⎝2µ11 0 0 · · · 0 0 0 · · · 0

0 µ12 0 · · · 0 0 0 · · · 0...

......

......

......

......

0 0 0 · · · 0 µ1(n−1) 0 · · · 0

⎞⎟⎟⎟⎠ ,

...

D(n−κ0)×|S0| =

⎛⎜⎜⎜⎝0 · · · 0 2µκ0κ0 0 0 · · · 0 00 · · · 0 0 µκ0(κ0+1) 0 · · · 0 0...

......

......

......

......

0 · · · 0 0 0 0 · · · 0 µκ0(n−1)

⎞⎟⎟⎟⎠ .

Hence f(ζ, 0) = ζ, φkl(ζ, 0) = µk+µl

µklζkζl for k ≤ κ0 and k < l < n, φkk(ζ, 0) = µk

µkkζk

2

for k ≤ κ0. Putting (z, w) = 0 and η = 0 in (11), we get g(ζ, 0) = 0. Sinceµkl =

√µk + µl, µkk =

√µk for k < l < n and k ≤ κ0 (see Theorem 2.2), from the

above argument, we have proved the following.

f(z, 0) = z;φkl(z, 0) =

√µk + µlzkzl, 1 ≤ k ≤ k0, k < l ≤ n− 1;

φkk(z, 0) =√µkz

2k, 1 ≤ k ≤ κ0;

g(z, 0) = 0.

(12)

Step 2. Determine F (z, w). We claim:

fj = zj+(b+ i4 )zjw

1+(b− i4 )w

, 1 ≤ j ≤ κ0;fj = zj , κ0 < j ≤ n− 1;φjl =

√2zjzl

1+(b− i4 )w

, 1 ≤ j < l ≤ κ0;φjl = zjzl

1+(b− i4 )w

, 1 ≤ j ≤ κ0, κ0 + 1 ≤ l ≤ n− 1;

φjj = z2j

1+(b− i4 )w

, 1 ≤ j ≤ κ0;g = w,

(13)

where b ∈ R is a real number.In fact, Since deg(F ) = 2, by (12), we can write F in the form

242 S. JI AND D. XU

fj =A

(1)j (z)+A

(2)j (z)+

˜

A(1)j (z)w+A′

jw+A′′jw2

1+E(1)(z)+E(2)(z)+ ˜E(1)(z)w+e1w+e2w2, 1 ≤ j ≤ n− 1;

φjl =B

(1)jl (z)+B

(2)jl (z)+

˜

B(1)jl (z)w+B′

jlw+B′′jlw

2

1+E(1)(z)+E(2)(z)+ ˜E(1)(z)w+e1w+e2w2, (j, l) ∈ S0;

g = C(1)(z)+C(2)(z)+ ˜C(1)(z)w+C′w+C′′w2

1+E(1)(z)+E(2)(z)+ ˜E(1)(z)w+e1w+e2w2.

(14)

Here we use notation h(k)(z) to denote a homogeneous polynomial of z with totaldegree k. We write φjl as a Taylor series at 0 and compare the expression with (12).Then we get

B(2)jl (z) = µjlzjzl, B

(1)jl (z) = B′

jl = B′′jl = ˜

B(1)jl (z) = 0, ∀(j, l) ∈ S0.

Applying (12) to φjl(z, 0), we obtain E(1)(z) = E(2)(z) = 0. Similarly, writing fj asa Taylor series at 0 and compare it with (3), we have A(1)

j (z) = zj , A′j = A′′

j = 0,˜

A(1)j (z) = i

2µjzj + e1zj for j ≤ κ0, and ˜

A(1)j (z) = e1zj for κ0 + 1 ≤ j ≤ n − 1. By

using (12) and the fact that E(1)(z) = E(2)(z) = 0 to fj(z, 0), we get A(2)j (z) = 0.

For g, we similarly obtain C ′ = 1, C(1)(z) = ˜C(1)(z) = C(2)(z) = 0. Sincedeg(F ) ≤ 2, as the proof of [HJ 2001, Lemma 6.1], using the last two equations of

(3), we get g(z, w) ≡ w. Therefore from (14), we find C ′ = 1, C ′′ = e1,˜E(1)(z) = 0

and e2 = 0. Combining the above results, we get

fj = zj+( i2 µjzj+e1zj)w

1+e1w , 1 ≤ j ≤ κ0; fj = zj+e1zjw1+e1w , κ0 < j ≤ n− 1;

φjl = µjlzjzl

1+e1w , ∀(j, l) ∈ S0; g = w.(15)

Since F maps ∂Hn into ∂HN, we have Im(g) = |f |2 + |φ|2 on ∂Hn. Noticeg(z, w) = w, this equation can be written into

|z|2 = |f(z, w)|2 + |φ(z, w)|2, ∀(z, w) ∈ ∂Hn. (16)

Replacing f, φ by the ones in (15), we can write (16) as

|z|2|1 + e1w|2 =κ0∑

j=1

∣∣∣∣zj + (i

2µjzj + e1zj)w

∣∣∣∣2 +n−1∑

j=κ0+1

|zj + e1zjw|2 +∑

(jl)∈S0

|µjlzjzl|2

for any (z, w) ∈ ∂Hn. Since w = u+ i|z|2, we obtain several equations:

|z|2|1 + e1(u+ i|z|2)|2 = |z|2(

1 + e1u+ e1u+ ie1|z|2 − ie1|z|2 + |e1|2u2 + |e1|2|z|4),

κ0∑j=1

∣∣∣∣zj + (i

2µjzj + e1zj)(u+ i|z|2)

∣∣∣∣2

=κ0∑

j=1

|zj |2(

1 + e1u+ e1u+ ie1|z|2 − ie1|z|2 + |e1|2u2 + |e1|2|z|4 − µj |z|2)

+κ0∑

j=1

|zj |2(

14u2µ2

j +i

2µje1u

2 − i

2µje1u

2 +14µ2

j |z|4 +i

2µje1|z|4 − i

2µje1|z|4

),


n−1∑j=κ0+1

|zj + e1zj(u+ i|z|2)|2

=n−1∑

j=κ0+1

|zj |2(

1 + e1u+ e1u+ ie1|z|2 − ie1|z|2 + |e1|2u2 + |e1|2|z|4),

and

|φ|2 =∑

(j,l)∈S0

µ2jl|zj |2|zl|2.

Substituting the above terms into (16), we get

0 =−κ0∑

j=1

µj |zj |2|z|2 + (u2 + |z|4)(

14

κ0∑j=1

|zj |2µ2j +

i

2e1

κ0∑j=1

µj |zj |2 − i

2e1

κ0∑j=1

µj |zj |2)

+∑

(jl)∈S0

µ2j,l|zj |2|zl|2, ∀z ∈ Cn−1 and u ∈ R.

Since z, u are independent variables,

14

κ0∑j=1

|zj |2µ2j +

i

2e1

κ0∑j=1

µj |zj |2 − i

2e1

κ0∑j=1

µj |zj |2 ≡ 0.

This meansκ0∑

j=1

(14µ2

j +i

2µje1 − i

2µje1

)|zj |2 ≡ 0.

Therefore 14µ

2j + i

2µje1 − i2µje1 = 0, ∀j = 1, · · · , κ0. Since µ1 = 1 (see Theorem 2.2),

this implies Im(e1) = − 14 and 1 = µ1 = µ2 · · · = µκ0 . Our claim (13) is proved.

Step 3. F is equivalent to Whitney Map. Let F be of the form (13).F = (f, φ, g) is equivalent to

F ′ =(λf(

z

λ,w

λ2), λφ(

z

λ,w

λ2), λ2g(

z

λ,w

λ2)).

Let λ = 12 . By permutating the components of F ′, we may assume that F ′ is of the

following form F ′(z, w) = (ψ1, · · · , ψκ0+1), where

ψ1 =(

2z21

1 + (4b− i)w,

2√

2z1z21 + (4b− i)w

, · · · , 2√

2z1zκ0

1 + (4b− i)w,

2z1zκ0+1

1 + (4b− i)w, · · · ,

2z1zn−1

1 + (4b− i)w,z1 + (4b+ i)z1w

1 + (4b− i)w

),

ψ2 =(

2z22

1 + (4b− i)w,

2√

2z2z31 + (4b− i)w

, · · · , 2√

2z2zκ0

1 + (4b− i)w,

2z2zκ0+1

1 + (4b− i)w, · · · ,

2z2zn−1

1 + (4b− i)w,z2 + (4b+ i)z2w

1 + (4b− i)w

),

244 S. JI AND D. XU

...

ψκ0 =(

2z2κ0

1 + (4b− i)w,

2zκ0zκ0+1

1 + (4b− i)w, · · · , 2zκ0zn−1

1 + (4b− i)w,zκ0 + (4b+ i)zκ0w

1 + (4b− i)w

),

ψκ0+1 = (zκ0+1, · · · , zn−1, w).

Using the Cayley transformations, F ′ induces a proper holomorphic mappingF = ρN ◦ F ′ ◦ ρ−1

n in Rat(Bn,Bn) given by F = (ψ1, · · · , ˜ψκ0+1), where

ψ1 =(

z21

1 + 2bi− 2biw,

√2z1z2

1 + 2bi− 2biw, · · · ,

√2z1zκ0

1 + 2bi− 2biw,

z1zκ0+1

1 + 2bi− 2biw, · · · ,

z1zn−1

1 + 2bi− 2biw,z1w(1 − 2bi) + 2biz1

1 + 2bi− 2biw

),

ψ2 =(

z22

1 + 2bi− 2biw,

√2z2z3

1 + 2bi− 2biw, · · · ,

√2z2zκ0

1 + 2bi− 2biw,

z2zκ0+1

1 + 2bi− 2biw, · · · ,

z2zn−1

1 + 2bi− 2biw,z2w(1 − 2bi) + 2biz2

1 + 2bi− 2biw

),

...

ψκ0 =(

z2κ0

1 + 2bi− 2biw,

zκ0zκ0+1

1 + 2bi− 2biw, · · · , zκ0zn−1

1 + 2bi− 2biw,zκ0w(1 − 2bi) + 2bizκ0

1 + 2bi− 2biw

),

˜ψκ0+1 = (zκ0+1, · · · , zn−1, w).

Consider

σ(z, w) =(z,i(1 + 2ib)√

1 + 4b2w

), and τ(z∗, w∗) =

(1 + 2ib√1 + 4b2

z∗,(1 + 2ib)i√

1 + 4b2w∗),

which are elements in Aut(Bn) and Aut(BN ) respectively. By definition, F is equiv-

alent to F ′ = τ ◦ F ◦σ. Replacing b by −b, as before, we write F ′ = (ψ1

′, · · · , ˜ψκ0+1

′)

where

ψ1

′=(

z21√

1 + 4b2(1 − 2b√1+4b2

w),

√2z1z2√

1 + 4b2(1 − 2b√1+4b2

w), · · · ,

√2z1zκ0√

1 + 4b2(1 − 2b√1+4b2

w),

z1zκ0+1√1 + 4b2(1 − 2b√

1+4b2w), · · · , z1zn−1√

1 + 4b2(1 − 2b√1+4b2

w),−iz1

2b√1+4b2

− w

1 − 2b√1+4b2

w

),

ψ2

′=(

z22√

1 + 4b2(1 − 2b√1+4b2

w),

√2z2z3√

1 + 4b2(1 − 2b√1+4b2

w), · · · ,

√2z2zκ0√

1 + 4b2(1 − 2b√1+4b2

w),

z2zκ0+1√1 + 4b2(1 − 2b√

1+4b2w), · · · , z2zn−1√

1 + 4b2(1 − 2b√1+4b2

w),−iz2

2b√1+4b2

− w

1 − 2b√1+4b2

w

),


...

ψκ0

′=(

z2κ0√

1 + 4b2(1 − 2b√1+4b2

w),

zκ0zκ0+1√1 + 4b2(1 − 2b√

1+4b2w), · · · ,

zκ0zn−1√1 + 4b2(1 − 2b√

1+4b2w),−izκ0

2b√1+4b2

− w

1 − 2b√1+4b2

w

),

˜ψκ0+1

′=(

1 − 2ib√1 + 4b2

zκ0+1, · · · , 1 − 2ib√1 + 4b2

zn−1,−(

1 − 2ib√1 + 4b2

)2

w

).

Define β = 2b√1+4b2

and Sβ =√

1 − β2. Noticing | 1−2ib√1+4b2

| = 1, we multiply i to

the last components of ψ1

′, · · · , ψκ0

′, −(√

1+4b2

1−2ib

)2

to the last component of ˜ψκ0+1

′

and√

1+4b2

1−2ib to the other components of ˜ψκ0+1

′, F ′ is equivalent to a new map, still

denote it as F ′ = (ψ1

′, · · · , ˜ψκ0+1

′):

ψ1

′=(

Sβz21

1−βw ,√

2Sβz1z21−βw , · · · ,

√2Sβz1zκ01−βw ,

Sβz1zκ0+1

1−βw , · · · , Sβz1zn−11−βw , z1

β−w1−βw

),

ψ2

′=(

Sβz22

1−βw ,√

2Sβz2z31−βw , · · · ,

√2Sβz2zκ01−βw ,

Sβz2zκ0+1

1−βw , · · · , Sβz2zn−11−βw , z2

β−w1−βw

),

...

ψκ0

′=(

Sβz2κ0

1−βw ,Sβzκ0zκ0+1

1−βw , · · · , Sβzκ0zn−1

1−βw , zκ0β−w1−βw

),

˜ψκ0+1

′= (zκ0+1, · · · , zn−1, w).

(17)

On the other hand, for any a ∈ R, we define

ϕa :=(

Saz

1 − aw,a− w

1 − aw

), Sa =

√1 − a2,

which is an automorphism of Bn. Similarly, we can define ϕ∗a ∈ Aut(BN ). For

Whitney map Wn,κ0 = (Γ1, · · · ,Γκ0+1) from Bn to BN defined in (9), we see thatϕ∗

a ◦Wn,κ0 ◦ ϕa has the following form

Γ1 =(

Saz21

1−aw ,√

2Saz1z21−aw , · · · ,

√2Saz1zκ01−aw ,

Saz1zκ0+1

1−aw , · · · , Saz1zn−11−aw , z1

a−w1−aw

),

Γ2 =(

Saz22

1−aw ,√

2Saz2z31−aw , · · · ,

√2Saz2zκ01−aw ,

Saz2zκ0+1

1−aw , · · · , Saz2zn−11−aw , z2

a−w1−aw

),

...

Γκ0 =(

Saz2κ0

1−aw ,Sazκ0zκ0+1

1−aw , · · · , Sazκ0zn−1

1−aw , zκ0a−w1−aw

),

˜Γκ0+1 = (zκ0+1, · · · , zn−1, w).

(18)

246 S. JI AND D. XU

Comparing (17) with (18), if we put a = 2b√1+4b2

, F ′ = ϕ∗a ◦Wn,κ0 ◦ϕa and this proves

Theorem 4.1.

5. Determining F (z, 0). From now on, we always consider F ∈ Rat(∂H4, ∂H9)with geometric rank 2 and degree 2 as in Theorem 1.2. In order to determine F (z, w),we need to determine F (z, 0) first. Let us denote φjl(z, w) =

∑u,v,s,t b

(jl)uvstz

u1 z

v2z

s3w

t.

Lemma 5.1. Let F ∈ Rat(∂H4, ∂H9) satisfying (3) with κ0 = 2 and deg(F ) = 2.Then

f1(z, 0) = z1, f2(z, 0) = z2, f3(z, 0) = z3,

φ11(z, 0) =z21

1 − 2ib(11)1001z1 − 2i√

1 + µ2b(12)1001z2 − 2ib(13)1001z3

,

φ12(z, 0) =√

1 + µ2z1z2

1 − 2ib(11)1001z1 − 2i√

1 + µ2b(12)1001z2 − 2ib(13)1001z3

,

φ13(z, 0) =z1z3

1 − 2ib(11)1001z1 − 2i√

1 + µ2b(12)1001z2 − 2ib(13)1001z3

,

φ22(z, 0) =√µ2z

22

1 − 2ib(11)1001z1 − 2i√

1 + µ2b(12)1001z2 − 2ib(13)1001z3

,

φ23(z, 0) =√µ2z2z3

1 − 2ib(11)1001z1 − 2i√

1 + µ2b(12)1001z2 − 2ib(13)1001z3

,

and b(22)1001 = b

(23)1001 = b

(11)0101 = b

(13)0101 = 0, b(12)0101 = µ2√

1+µ2b(11)1001, b

(23)0101 =

√µ2b

(13)1001,

b(22)0101 =

√µ2(1 + µ2)b

(12)1001, where µ2 ≥ 1.

Proof. As we did in §3 and by the same notation, we have

A=

⎛⎜⎜⎜⎜⎝−2ζ1 0 0−ζ2 −µ2ζ1 0−ζ3 0 00 −2µ2ζ2 00 −µ2ζ3 0

⎞⎟⎟⎟⎟⎠ , and Aζt=

⎛⎜⎜⎜⎜⎜⎝−2ζ1

2

−(1 + µ2)ζ1ζ2−ζ1ζ3

−2µ2ζ22

−µ2ζ2ζ3

⎞⎟⎟⎟⎟⎟⎠ .

We can write the 5 × 5 matrix B = D + B where

D =

⎛⎜⎜⎜⎜⎝2 0 0 0 00

√1 + µ2 0 0 0

0 0 1 0 00 0 0 2

√µ2 0

0 0 0 0√µ2

⎞⎟⎟⎟⎟⎠ ,

B =

⎛⎜⎜⎜⎜⎝4iζ1 02iζ2 2iζ12iζ3 00 4iζ20 2iζ3

⎞⎟⎟⎟⎟⎠(b(11)1001 b

(12)1001 b

(13)1001 b

(22)1001 b

(23)1001

b(11)0101 b

(12)0101 b

(13)0101 b

(22)0101 b

(23)0101

).


By (8), we have −B−1Aζt= −[I − 2i�ζ

(I + 2i�b · �ζ)−1 ·�b]D−1Aζ

t. Writing(

C11 C12

C21 C22

):= 2i�b · �ζ

= 2i

(b(11)1001 b

(12)1001 b

(13)1001 b

(22)1001 b

(23)1001

b(11)0101 b

(12)0101 b

(13)0101 b

(22)0101 b

(23)0101

)⎛⎜⎜⎜⎜⎜⎝ζ1 01√

1+µ2ζ2

1√1+µ2

ζ1

ζ3 00 1√

µ2ζ2

0 1√µ2ζ3

⎞⎟⎟⎟⎟⎟⎠= 2i

⎛⎝ b(11)1001ζ1 + b

(12)1001√1+µ2

ζ2 + b(13)1001ζ3

b(12)1001√1+µ2

ζ1 + b(22)1001√µ2ζ2 + b

(23)1001√µ2ζ3

b(11)0101ζ1 + b

(12)0101√1+µ2

ζ2 + b(13)0101ζ3

b(12)0101√1+µ2

ζ1 + b(22)0101√µ2ζ2 + b

(23)0101√µ2ζ3

⎞⎠ ,

we have

−B−1 ·A · ζ =[I − 2i

�ζ ·(

1 + C22 −C12

−C21 1 + C11

)·�b

det

(1 + C11 C12

C21 1 + C22

) ]⎛⎜⎜⎜⎜⎜⎝

ζ12

√1 + µ2ζ1ζ2ζ1ζ3√µ2ζ2

2

√µ2ζ2ζ3

⎞⎟⎟⎟⎟⎟⎠ .

Denoting by ∆ the determinant det(

1 + C11 C12

C21 1 + C22

)and by (7), by direct com-

putation, we obtain

∆(ζ, 0)φ11(ζ, 0) = ζ12

+ 2ib(12)0101√

1+µ2ζ1

3+(

2ib(12)1001√

1+µ2+ 2ib

(22)0101√µ2

− 2i√

1 + µ2b(12)1001

)ζ1

2ζ2

+ 2ib(23)0101√µ2

ζ12ζ3 − 2i

√µ2b

(22)1001ζ1ζ2

2 − 2i√µ2b

(23)1001ζ1ζ2ζ3.

(19)Since F is of degree 2, the numerator of φ11(ζ, 0) must be ζ

2

1 by (3) so that from (19)

we get b(22)1001 = b(23)1001 = 0 and then

φ11 =ζ1

2

∆

[1 +

2ib(12)0101√1 + µ2

ζ1 +(

2ib(12)1001√1 + µ2

+2ib(22)0101√

µ2− 2i

√1 + µ2b

(12)1001

)ζ2 +

2ib(23)0101√µ2

ζ3

].

(20)Similarly we calculate φ22 to obtain b(11)0101 = b

(13)0101 = 0 and at the value (ζ, 0)

φ22 =√µ2ζ2

2

∆

[1 + 2i

(b(11)1001 −

1µ2

√1 + µ2

b(12)0101

)ζ1 +

2ib(12)1001√1 + µ2

ζ2 + 2ib(13)1001ζ3

], (21)

φ13 =ζ1ζ3∆

[1 +

2ib(12)0101√1 + µ2

ζ1 +(

2ib(12)1001√1 + µ2

+2ib(22)0101√

µ2− 2i

√1 + µ2b

(12)1001

)ζ2 +

2ib(23)0101√µ2

ζ3

],

(22)

φ23 =√µ2ζ2ζ3

∆

[1 + 2i

(b(11)1001 −

1µ2

√1 + µ2

b(12)0101

)ζ1 +

2ib(12)1001√1 + µ2

ζ2 + 2ib(13)1001ζ3

], (23)

248 S. JI AND D. XU

φ12 =√

1 + µ2ζ1ζ2∆

[1+

2iµ2

1 + µ2b(11)1001ζ1+

2ib(22)0101√µ2(1 + µ2)

ζ2+2i(µ2b

(13)1001

1 + µ2+

√µ2b

(23)0101

µ2(1 + µ2)

)ζ3

].

(24)With the fact that b(22)1001 = b

(23)1001 = b

(11)0101 = b

(13)0101 = 0, we also calculate

∆(ζ, 0)= det(1 + 2i�b · �ζ) = 1 + C11 + C22 + C11C22 − C12C21

= 1 + ζ1

(2ib(11)1001 + 2i

b(12)0101√

1 + µ2

)+ ζ2

(2i

b(12)1001√

1 + µ2+ 2i

b(22)0101√µ2

)+ ζ3

(2ib(13)1001 + 2i

b(23)0101√µ2

)− 4ζ1

2 b(11)1001b

(12)0101√

1 + µ2− 4ζ2

2 b(12)1001b

(22)0101√

1 + µ2√µ2

− 4ζ32 b

(13)1001b

(23)0101√µ2

− 4ζ1ζ2b(11)1001b

(22)0101√µ2

+ ζ1ζ3

(− 4

b(11)1001b

(23)0101√µ2

− 4b(13)1001b

(12)0101√

1 + µ2

)+ ζ2ζ3

(− 4

b(12)1001b

(23)0101√

(1 + µ2)µ2

− 4b(13)1001b

(22)0101√µ2

).

The factors in the right hand sides of (20)(21)...(24) must be the same, and it alsomust be a factor of ∆(ζ, 0). Then Lemma 5.1 follows immediately.

6. Determining F (z, w).

Theorem 6.1. Let F : ∂H4 → ∂H9 satisfies (3) with κ0 = 2 and deg(F ) = 2.Then F must be of the form:

f1=z1 − 2ib(11)1001z

21 − 2ib(13)1001z1z3 + (E0001 + i

2 )z1w

1 − 2ib(11)1001z1 − 2ib(13)1001z3 + E0001w,

f2=z2 − 2ib(11)1001z1z2 − 2ib(13)1001z2z3 + (E0001 + iµ2

2 )z2w

1 − 2ib(11)1001z1 − 2ib(13)1001z3 + E0001w,

f3=z3 − 2ib(13)1001z

23 − 2ib(11)1001z1z3 + E0001z3w

1 − 2ib(11)1001z1 − 2ib(13)1001z3 + E0001w,

φ11=z21 + b

(11)1001z1w

1 − 2ib(11)1001z1 − 2ib(13)1001z3 + E0001w, φ12 =

√1 + µ2z1z2 + µ2√

1+µ2b(11)1001z2w

1 − 2ib(11)1001z1 − 2ib(13)1001z3 + E0001w,

φ13=z1z3 + b

(13)1001z1w

1 − 2ib(11)1001z1 − 2ib(13)1001z3 + E0001w, φ22 =

√µ2z

22

1 − 2ib(11)1001z1 − 2ib(13)1001z3 + E0001w,

φ23=√µ2z2z3 +

√µ2b

(13)1001z2w

1 − 2ib(11)1001z1 − 2ib(13)1001z3 + E0001w, g = w,


where b(11)1001, b(13)1001, E0001 ∈ C and µ2 ≥ 1 with

|b(11)1001|2 =µ2

2 − 14

, Im(E0001) = −µ22

4− |b(13)1001|2. (25)

Proof. By the normalization condition (3) and Lemma 5.1, we can write themap F in the following form:

f1=[z1 − 2ib(11)1001z

21 − 2i

√1 + µ2b

(12)1001z1z2 − 2ib(13)1001z1z3 + (E0001 +

i

2)z1w

]·[1 − 2ib(11)1001z1 − 2i

√1 + µ2b

(12)1001z2 − 2ib(13)1001z3 + E0001w + E1001z1w +

+E0101z2w + E0011z3w + E0002w2

]−1

,

f2=[z2 − 2i

√1 + µb

(12)1001z

22 − 2ib(11)1001z1z2 − 2ib(13)1001z2z3 + (E0001 +

iµ2

2)z2w

]·[1 − 2ib(11)1001z1 − 2i

√1 + µ2b

(12)1001z2 − 2ib(13)1001z3 + E0001w + E1001z1w +

+E0101z2w + E0011z3w + E0002w2

]−1

,

f3=[z3 − 2ib(13)1001z

23 − 2ib(11)1001z1z3 − 2i

√1 + µ2b

(12)1001z2z3 + E0001z3w

]·[1 − 2ib(11)1001z1 − 2i

√1 + µ2b

(12)1001z2 − 2ib(13)1001z3 + E0001w + E1001z1w +

+E0101z2w + E0011z3w + E0002w2

]−1

,

φ11=[z21 + b

(11)1001z1w

][1 − 2ib(11)1001z1 − 2i

√1 + µ2b

(12)1001z2

−2ib(13)1001z3 + E0001w + E1001z1w + E0101z2w + E0011z3w + E0002w2

]−1

,

φ12=[√

1 + µ2z1z2 + b(12)1001z1w +

µ2√1 + µ2

b(11)1001z2w

][1 − 2ib(11)1001z1 − 2i

√1 + µ2b

(12)1001z2


]−1

,

φ13=[z1z3 + b

(13)1001z1w

][1 − 2ib(11)1001z1 − 2i

√1 + µ2b

(12)1001z2


]−1

,

250 S. JI AND D. XU

φ22=[√

µ2z22 +

√µ2(1 + µ2)b

(12)1001z2w

][1 − 2ib(11)1001z1 − 2i

√1 + µ2b

(12)1001z2


]−1

,

φ23=[√

µ2z2z3 +√µ2b

(13)1001z2w

][1 − 2ib(11)1001z1 − 2i

√1 + µ2b

(12)1001z2


]−1

,

g=[w + C1001z1w + C0101z2w + C0011z3w + E0001w

2

][1 − 2ib(11)1001z1 − 2i

√1 + µ2b

(12)1001z2


]−1

.

When z2 = z3 = 0 and Im(w) = z1, we get f2 = f3 = φ22 = φ23 = 0, and

f1 =z1 − 2ib(11)1001z

21 + (E0001 + i

2 )z1w

1 − 2ib(11)1001z1 + E0001w + E1001z1w + E0002w2,

φ11=z21 + b

(11)1001z1w

1 − 2ib(11)1001z1 + E0001w + E1001z1w + E0002w2,

φ12=b(12)1001z1w

1 − 2ib(11)1001z1 + E0001w + E1001z1w + E0002w2,

φ13=b(13)1001z1w

1 − 2ib(11)1001z1 + E0001w + E1001z1w + E0002w2,

g =w + C1001z1w + E0001w

2

1 − 2ib(11)1001z1 + E0001w + E1001z1w + E0002w2.

Consider the basic equation Im(g) = |f |2 + |φ|2 for any Im(w) = |z1|2. Byconsidering the u2 terms, we see that E0002 is real. Considering the u4 terms, we

obtain E0002(E0001 − E0001) = 0. Considering the uz1 terms, C1001 + 2ib(11)1001 = 0.Considering the u2z1 terms, we get E1001 = 0. Considering the u3z1 terms, weget C1001E0002 = 0. Similarly, when z1 = z3 = 0 and Im(w) = |z2|2, we get

C0101 = −2i√

1 + µ2b(12)1001, E0101 = 0, C0101E0002 = 0; when z1 = z2 = 0 and

Im(w) = |z3|2, we get C0011 = −2ib(13)1001, E0011 = 0, C0011E0002 = 0. Therefore wecan distinguish two cases.


Case i: C1001 = C0101 = C0011 = 0. Then b(11)1001 = b

(12)1001 = b

(13)1001 = 0 and we

apply Claim (13) in the proof of Theorem 4.1 to know that F is of the form (13). Weare done.

Case ii: (C1001, C0101, C0011) = (0, 0, 0). In this case E0002 = 0. By the basicequation Im(g) = |f |2 on ∂H4, we have

|z|2|1 − 2ib(11)1001z1 − 2i√

1 + µ2b(12)1001z2 − 2ib(13)1001z3 +E0001w|2

= |z1|2|1 − 2ib(11)1001z1 − 2i√

1 + µ2b(12)1001z2 − 2ib(13)1001z3 + (E0001 +

i

2)w|2

+|z2|2|1 − 2i√

1 + µb(12)1001z2 − 2ib(11)1001z1 − 2ib(13)1001z3 + (E0001 +

iµ2

2)w|2

+|z3|2|1 − 2ib(13)1001z3 − 2ib(11)1001z1 − 2i√

1 + µ2b(12)1001z2 + E0001w|2

+|z1|2|z1 + b(11)1001w|2 + |

√1 + µ2z1z2 + b

(12)1001z1w +

µ2√1 + µ2

b(11)1001z2w|2

+|z1|2|z3 + b(13)1001w|2 + |z2|2|√µ2z2 +

√µ2(1 + µ2)b

(12)1001w|2

+|z2|2|√µ2z3 +√µ2b

(13)1001w|2.

By considering the z1z2u2 terms, we get b(12)1001b(11)1001 = 0. We consider z2 = z3 = 0 and

Im(w) = |z1|2, divided by |z1|2, to get

|1 − 2ib(11)1001z1 + E0001w|2

= |1 − 2ib(11)1001z1 + (E0001 +i

2)w|2 + |z1 + b

(11)1001w|2 + |b(12)1001w|2 + |b(13)1001w|2.

By considering the u2 terms, we have

Im(E0001) = −14− |b(11)1001|2 − |b(12)1001|2 − |b(13)1001|2. (26)

We also consider z1 = z3 = 0 and Im(w) = |z2|2, divided by the |z2|2 terms, to get:

|1 − 2i√

1 + µ2b(12)1001z2 + E0001w|2 = |1 − 2i

√1 + µb

(12)1001z2 + (E0001 +

iµ2

2)w|2

+| µ2√1 + µ2

b(11)1001w|2 + |√µ2z2 +

√µ2(1 + µ2)b

(12)1001w|2 + |√µ2b

(13)1001w|2.

By considering the u2 terms, divided by µ2, we have Im(E0001) = −µ24 − µ2

1+µ2|b(11)1001|2−

(1+µ2)|b(12)1001|2−|b(13)1001|2. From above two formulas, we get 0 = µ2−14 − 1

1+µ2|b(11)1001|2 +

µ2|b(12)1001|2 so that |b(11)1001|2 = µ22−14 + µ2(1 + µ2)|b(12)1001|2. Recalling b(12)1001b

(11)1001 = 0 and

µ2 ≥ 1, we obtain the equation |b(12)1001|2(

µ22−14 + µ2(1 + µ2)|b(12)1001|2

)= 0. Then either

b(12)1001 = 0 or µ2

2−14 +µ2(1+µ2)|b(12)1001|2 = 0. Since µ2 ≥ 1, the second possibility implies

b(12)1001 = µ2 = 0. We have proved b

(12)1001 = 0 and |b(11)1001|2 = µ2

2−14 . Therefore our basic

252 S. JI AND D. XU

equation becomes

|z|2|1 − 2ib(11)1001z1 − 2ib(13)1001z3 + E0001w|2

= |z1|2|1 − 2ib(11)1001z1 − 2ib(13)1001z3 + (E0001 +i

2)w|2

+|z2|2|1 − 2ib(11)1001z1 − 2ib(13)1001z3 + (E0001 +iµ2

2)w|2

+|z3|2|1 − 2ib(13)1001z3 − 2ib(11)1001z1 + E0001w|2 + |z1|2|z1 + b(11)1001w|2

+|z2|2|√

1 + µ2z1 +µ2√

1 + µ2b(11)1001w|2 + |z1|2|z3 + b

(13)1001w|2

+|z2|2|√µ2z2|2 + |z2|2|√µ2z3 +√µ2b

(13)1001w|2.

Considering all terms involving u2, we get

|z|2|E0001|2 = |z1|2|E0001 +i

2|2 + |z2|2|E0001 +

iµ2

2|2 + |z3|2|E0001|2

+|z1|2|b(11)1001|2 + |z2|2| µ2√1 + µ2

b(11)1001|2 + |z1|2|b(13)1001|2 + |z2|2|√µ2b

(13)1001|2.

By recalling |b(11)1001|2 = µ22−14 , we have

0 = |z1|2(

14

+ Im(E0001))

+ |z2|2(µ2

2

4+ µ2Im(E0001)

)+µ2

2 − 14

|z1|2 +µ2

2(µ2 − 1)4

|z2|2 +(|z1|2 + µ2|z2|2

)|b(13)1001|2.

Comparing the |z1|2 and |z2|2 respectively, we find out

0 =(

14

+ Im(E0001))

+µ2

2 − 14

+ |b(13)1001|2,

0 =(µ2

4+ Im(E0001)

)+µ2(µ2 − 1)

4+ |b(13)1001|2.

Thus we have proved that |b(11)1001|2 = µ22−14 and Im(E0001) = −µ2

24 − |b(13)1001|2.

Corollary 6.2. Let F ∈ Rat(H4,H9) be as in Theorem 6.1. Then F is equiv-alent to another map that is of the form as in Theorem 6.1 with the same µ2 valueand satisfies the additional property:

b(13)1001 = 0 and Re(E0001) = 0. (27)

Proof. Let F be as in Theorem 6.1. We take σ ∈ Aut0(H4) and τ∗ ∈ Aut0(H9)

in the forms of (4) and (5) with U = Id, λ = 1, U∗22 = Id, r = 0 and a =

⎛⎝ 00

−b(13)1001

⎞⎠.

Then by Lemma 2.4, we can verify that F = τ∗ ◦ F ◦ σ still is of the form as inTheorem 6.1, with the same µ2 value, and satisfies b(13)1001 = 0.

Next fixing F that is of the form as in Theorem 6.1 with b(13)1001 = 0. We again

take σ ∈ Aut0(H4) and τ∗ ∈ Aut0(H9) in the forms of (4) and (5) with U = Id, λ =


1, a = 0, U∗22 = Id and r = −Re(E0001). Then by Lemma 2.4, we can verify that

F = τ∗ ◦F ◦ σ is of the form as in Theorem 6.1, with the same µ2 value, and satisfies(27).

7. The Proof of Theorem 1.2. Let F ∈ Rat(H4,H9) be as in Corollary 6.2.For any p ∈ ∂H4, let µ1,F (p) and µ2,F (p), with µ1,F (p) ≤ µ2,F (p), be the eigenvalues

of the semipositive matrix A(p) : = (ajl(p)) = (−2i ∂2f∗∗j,p

∂zl∂w |0). Define µ1,F (p) = 1 and

µ2,F (p) = µ2,F (p)µ1,F (p) ≥ 1. Recall that µ1,F (p) and µ2,F (p) are the coefficients µ1 = 1 and

µ2 in Theorem 2.2, respectively, which are the eigenvalues of the semipositive matrix

(−2i∂2f∗∗∗j,p

∂zl∂w |0). Write p = (p1, ..., p7) = (z1, z1, z2, z2, z3, z3, u) that is identified as a

point in ∂H4, and write akk(p) = a(kk)0 +

∑7j=1 a

(kk)j pj + o(|p|), k = 1, 2, 3.

Lemma 7.1. Let F be as in Corollary 6.2. for any p that is closed to 0

a11(p) = 1 − 2ReE0001u+ 4i(b(11)1001z1 − b(11)1001z1) + o(|p|),

a22(p) = µ2 − 2µ2ReE0001u+ 2iµ2(b(11)1001z1 − b

(11)1001z1) + o(|p|),

a33(p) = o(|p|).

We shall assume this lemma, which will be proved in the next section, to proveTheorem 1.2. We first study the real analytic function µ2,F (p) of p.

Lemma 7.2. Let F be as in Corollary 6.2. Then

µ2,F (p) = µ2 + 4µ2Im

(b(11)1001z1

)+ o(|p|)

for any such p near 0, where µ2 is the one as in (25).

Proof. Recall that µ1,F (p) and µ2,F (p) must be the eigenvalues of the equationdet(λI −A(p)) = 0, i.e.,

det

⎡⎣ λ− a11(p) −a12(p) −a13(p)−a21(p) λ− a22(p) −a23(p)−a31(p) −a32(p) λ− a33(p)

⎤⎦ = 0.

Since the geometric rank is 2, we must have det(λI − A(p)) = λ(λ − µ1,F (p))(λ −µ2,F (p)). For simplicity, we denote aij = aij(p). Then

det(λI −A(p)) = λ

[λ2 − (a11 + a22 + a33)λ+ (a11a22 + a11a33

+a22a33 − a21a12 − a23a32 − a31a13)],

µ1,F (p) =12

[(a11 + a22 + a33) −

((a11 + a22 + a33)2 − 4(a11a22

+a11a33 + a22a33 − a21a12 − a23a32 − a31a13))1/2]

,

254 S. JI AND D. XU

µ2,F (p) =12

[(a11 + a22 + a33) +

((a11 + a22 + a33)2 − 4(a11a22

+a11a33 + a22a33 − a21a12 − a23a32 − a31a13))1/2]

.

Then

µ2,F (p) =µ2(p)µ1(p)

=1 +

√1 −N(p) +M(p)

1 −√1 −N(p) +M(p)

where, by Lemma 7.1,

N(p)= 4a11a22

(a11 + a22)2+ o(|p|), M(p) = 4

a21a12 + a13a31 + a23a32

(a11 + a22 + a33)2= o(|p|).

By considering the Tayler seriese of ajj = a(jj)0 +

∑k a

(jj)k pk + o(|p|) for k = 1, 2, 3,

we obtain

µ2,F (p) =a(22)0

a(11)0

− 1

(a(11)0 )2

7∑j=1

(a(22)0 a

(11)j − a

(11)0 a

(22)j

)pj + o(|p|).

Then from Lemma 7.1, the desired equality is proved.

Proof of Theorem 1.2. Let F be as in Corollary 6.2 and fixed . If its µ2 = 1,then F must be Whitney map (10). Suppose that µ2 > 1. Then by Lemma 7.2,we can choose p = (−ib(11)1001r, 0, 0, 0, 0, 0, 0) or identify p = p(r) = (−ib(11)1001r, 0,0,i|b(11)1001|2r) ∈ ∂H4, where r > 0, to conclude that

µ2,F (p(r)) = µ2 − 4µ2|b(11)1001|2r + o(|r|).

Therefore, there is a constant σ > 0 such that for any 0 < r < σ, the derivativedµ2,F (p(r))

dr < 0. Therefore for such p, µ2,Fp(r)(0) is decreasing as r increases. Henceµ2,Fp(r)(0) < µ2,F (0). In other words, we find a new map that has smaller µ2 value.By Corollary 6.2, we can assume that this new map is of the form as in Theorem 6.1,with the same µ2 value, b(13)1001 = 0 and Re(E0001) = 0. Let us denote this map asF1. Repeating this process, we obtain a sequence of maps {Fk}∞k=1 such that eachmap Fk is of the form as in Theorem 6.1 with b(13)1001 = 0 and Re(E0001) = 0 and thateach Fk is equivalent to the F and that µ2,Fk+1 < µ2,Fk

holds for all k. Then thelimit map F = limk Fk must be of the form as in Theorem 6.1 with b

(13)1001 = 0 and

Re(E0001) = 0, and with the minimum µ2,F value.We want to prove that this map is the desired one. In fact, suppose that µ2,F > 1.

Then p = 0 must be a critical point of the real analytic function µ2,F (p). By Lemma

7.2, p = 0 is a critical point if and only if b(11)1001 = 0. This implies, by (25), thatµ2,F = 1, which is a contradiction to the assumption that µ2,F > 1.

Finally, since we can assume that the map F is of the form as in Corollary 6.2with µ2 = 1, it is the Whiteny map (10).

To finish the proof of Theorem 1.2, it remains to prove Lemma 7.1.


8. The Proof of Lemma 7.1. We are going to use the following formula (seecf. [H 2003, § 2]) to prove Lemma 7.1:

ajl(p) = −2ip2f∗∗p,j

∂zl∂w|0 = −2i

{1

λ(p)LjT f(p) · Llf(p)

t

(28)

− 2iλ(p)2

(T f(p) · Llf(p)

t)(

T f(p) · Lj f(p)t)− δjl

2λ(p)

(T 2g(p) − 2iT 2f(p) · f(p)

t)}

.

Since F is of the form as in Corollary 6.2, we have

f1=z1 − 2ib(11)1001z

21 + (E0001 + i

2 )z1w

1 − 2ib(11)1001z1 + E0001w= z1 + o(|(z, w)|).

Then

Tf1 =(E0001 + i

2 )z1

1 − 2ib(11)1001z1 + E0001w− E(z1 − 2ib(11)1001z

21 + (E0001 + i

2 )z1w)

(1 − 2ib(11)1001z1 + E0001w)2

=i

2z1 + o(|(z, w)|),

Similarly, by direct computation, we obtain the following results. For simplicity, weuse o(1) to denote o(|(z, w)|).

T 2f1 = −iE0001z1 + o(1), L1f1 = 1 +i

2w + o(1),

L1Tf1 =i

2− 2b(11)1001z1 − iE0001w + o(1), L2f1 = o(1), L2Tf1 = o(1), L3f1 = o(1),

f2 = z2 + o(1), T f2 =iµ2

2z2 + o(1), T 2f2 = −iµ2E0001z2 + o(1),

L1f2 = o(1), L1Tf2 = −µ2b(11)1001z2 + o(1), L2f2 = 1 +

iµ2

2w + o(1),

L2Tf2 =iµ2

2− µ2b

(11)1001z1 − iµ2E0001w + o(1), L3f2 = o(1)

f3 = z3 + o(1), T f3 = o(1), T 2f3 = o(1), L1f3 = o(1), L1Tf3 = o(1),L2f3 = 0, L2Tf3 = 0, L3f3 = 1 + o(1), L3Tf3 = o(1),

φ11 = o(1), Tφ11 = b(11)1001z1 + o(1),

T 2φ11 = −2E0001b(11)1001z1 + o(1), L1φ11 = 2z1 + b

(11)1001w + o(1),

L1Tφ11 = b(11)1001 + (−2E0001 + 4i|b(11)1001|2)z1 − 2b(11)1001E0001w + o(1),

L2φ11 = 0, L2Tφ11 = 0, L3φ11 = o(1),

256 S. JI AND D. XU

φ12 = o(1), Tφ12 =µ2√

1 + µ2b(11)1001z2 + o(1),

T 2φ12 = −2E0001µ2√

1 + µ2b(11)1001z2 + o(1), L1φ12 =

√1 + µ2z2 + o(1),

L1Tφ12 = −E0001

√1 + µ2z2 + 2i|b(11)1001|2

µ2√1 + µ2

z2 + o(1),

L2φ12 =√

1 + µ2z1 +µ2√

1 + µ2b(11)1001w + o(1),

L2Tφ12 =µ2b

(11)1001√

1 + µ2+ (

2iµ2√1 + µ2

|b(11)1001|2 − E0001

√1 + µ2)z1 − 2µ2E0001b

(11)1001√

1 + µ2w + o(1),

L3φ12 = o(1),

φ13 = o(1), Tφ13 = o(1), T 2φ13 = o(1), L1φ13 = z3 + o(1), L1Tφ13 = o(1),L2φ13 = 0, L2Tφ13 = 0, L3φ13 = z1 + o(1), L3Tφ13 = −E0001z1 + o(1),

φ22 = o(1), Tφ22 = o(1), T 2φ22 = o(1),L1φ22 = o(1), L1Tφ22 = o(1), L2φ22 = 2

√µ2z2 + o(1),

L2Tφ22 = −2E0001√µ2z2 + o(1), L3φ22 = o(1),

φ23 = o(1), Tφ23 = o(1), T 2φ23 = o(1), L1φ23 = o(1),L1Tφ23 = o(1), L2φ23 =

√µ2z3 + o(1), L2Tφ23 = o(1),

L3φ23 =√µ2z2, L3Tφ23 = −E0001

√µ2z2 + o(1).

By (29) and all above formulas, the desired formulas in Lemma 7.1 are obtained.

REFERENCES

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Documents

1. Introduction. B C